std2-std1-common

Financial Maths, STD1 F3 2025 HSC 27

The graph shows the salvage value of a car over 5 years.

The salvage values are based on the declining-balance method.

By what amount will the car’s value depreciate during the 10th year? (4 marks)

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$$1476.40$

Show Worked Solution

$\text{Find}\ r:$

$\text{When}\ \ n=1, \ S=$44\ 000\ \ \text{(see graph)}$

$S$	$=V_0(1-r)^n$
$44\ 000$	$=55\ 000(1-r)^1$
$\dfrac{44\ 000}{55\ 000}$	$=1-r$
$1-r$	$=0.8$
$r$	$=1-0.8=0.20$

$\text{Find \(S$ when}\ \ n=9\ \ \text{and}\ \ n=10:\)

$S_9=55\ 000(1-0.20)^{9}=$7381.97504$

$S_{10}=55\ 000(1-0.20)^{10}=$5905.5800$

$S_9-S_{10}=$7381.9750-$5905.580=$1476.40\ \text{(nearest cent)}$

$\therefore\ \text{The car’s value will depreciate by \$1476.40 in the 10th year.}$

♦♦♦ Mean mark 23%.

Financial Maths, STD1 F3 2025 HSC 24

A used car has a sale price of $24 200. In addition to the sale price, the following costs are charged:

transfer of registration $50
stamp duty which is calculated at $3 for every $100, or part thereof, of the sale price.

Kat borrows the total amount to be paid for the car, including transfer of registration and stamp duty. Simple interest at the rate of 6.8% per annum is charged on the loan. The loan is to be repaid in equal monthly repayments over 3 years.

Calculate Kat’s monthly repayment. (5 marks)

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$$835.31$

Show Worked Solution

$\text{Stamp Duty} =\dfrac{24\ 200}{100}\times 3=$726$

$\text{Total Cost}\ $	$\text{= Price + Transfer + Stamp Duty}$
	$\text{= }24\ 200+50+726$
	$\text{= }$24\ 976$

$\text{Interest}=Prn=24\,976\times 0.068\times 3=$5095.104$

$\text{Loan amount}\ $	$\text{= Total Cost + Interest}$
	$\text{= }24\ 976+5095.104$
	$\text{= }$30\ 071.104$

$\text{3 years}= 3 \times 12=36\ \text{months}$

$\text{Monthly repayment}$	$=\dfrac{30\ 071.104}{36}$
	$=$835.308444$
	$\approx $835.31$

♦♦ Mean mark 46%.

Algebra, STD1 A3 2025 HSC 23

It costs $465 to register a passenger car and $350 to register a motorcycle.

Let	$P$	$\ =\ \text{the number of passenger cars, and}$
	$B$	$\ =\ \text{the number of motorcycles}$

Write TWO linear equations that represent the relationship below.

There are 11 times as many passenger cars as motorcycles.
The total registration fees for passenger cars and motorcycles is $494 million. (2 marks)

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$\text{Equation 1: }\ P=11B$

$\text{Equation 2: }\ 465P+350B=494\ 000\ 000$

Show Worked Solution

$\text{Equation 1: }\ P=11B$

$\text{Equation 2: }\ 465P+350B=494\ 000\ 000$

♦♦♦ Mean mark 11%.

Financial Maths, STD1 F3 2025 HSC 21

A credit card has an interest-free period of 45 days from and including the date of purchase. Interest is charged on purchases made, compounding daily at a rate of 13.74% per annum, from and including the day following the interest-free period.

Concert tickets were purchased for a total of $392 using this credit card.

Full payment was made on the 68th day from the date of purchase. There were no other purchases on this credit card.

What was the total interest charged when the account was paid in full? (3 marks)

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$\text{Interest charged}\ =\$ 3.41$

Show Worked Solution

$\text{Day 1-45: no interest is charged}$

$\text{Day 46-68: interest charged (23 days)}$

$\text{Daily interest rate}=\dfrac{13.74}{365} \%=\dfrac{13.74}{365 \times 100}$

$\text{Amount owing}=392\left(1+\dfrac{13.74}{365 \times 100}\right)^{23}=\$ 395.41$

$\text{Interest charged}=395.41-392=\$ 3.41$

♦♦ Mean mark 33%.

Measurement, STD1 M4 2025 HSC 17

Paint is sold in two sizes at a local shop.

Four-litre cans at $90 per can
Ten-litre cans at $205 per can

Mina needs to purchase 80 litres of paint.

Calculate the amount of money Mina will save by purchasing only ten-litre cans of paint rather than only four-litre cans of paint. (2 marks)

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$\text{Saving}=$160$

Show Worked Solution

$\text{4 litre cans needed}\ =\dfrac{80}{4}=20$

$\text{Cost using 4 litre cans}\ =20\times $90=$1800$

$\text{10 litre cans needed}\ =\dfrac{80}{10}=8$

$\text{Cost using 10 litre cans}\ =8\times $205=$1640$

$\therefore\ \text{Saving}\ =$1800-$1640=$160$

Measurement, STD1 M5 2025 HSC 10 MC

The ratio of the dimensions of a model car to the dimensions of an actual car is $1:64$. The actual car has a length of 4.9 m.

What is the length of the model car in cm, correct to 1 decimal place?

3.1
7.7
13.1
59.1

Show Answers Only

$B$

Show Worked Solution

$\text{Actual length}=4.9\ \text{m}\ =490\ \text{cm}$

$\therefore\ \text{Model car length}\ =490\times\dfrac{1}{64}=7.65625\approx 7.7\ \text{cm}$

$\Rightarrow B$

♦♦ Mean mark 38%.

Probability, STD1 S2 2025 HSC 8 MC

A spinner made up of 4 colours is spun 100 times. The frequency histogram shows the results.

Which of these spinners is most likely to give the results shown?

Show Answers Only

$A$

Show Worked Solution

$P(\text{White})$	$=\dfrac{50}{100}=\dfrac{1}{2}$
$P(\text{Red})$	$=\dfrac{25}{100}=\dfrac{1}{4}$
$P(\text{Yellow})$	$=\dfrac{15}{100}=\dfrac{3}{20}$
$P(\text{Green})$	$=\dfrac{10}{100}=\dfrac{2}{20}=\dfrac{1}{10}$

$\text{Eliminate Options B and D as white}\ \neq \dfrac{1}{2}\ \text{of spinner.}$

$\text{Eliminate Option C as red}\ \neq \dfrac{1}{4}\ \text{of spinner.}$

$\Rightarrow A$

Networks, STD1 N1 2025 HSC 6 MC

The network shows the distances, in kilometres, along a series of roads that connect towns.

What is the value of the largest weighted edge included in the minimum spanning tree for this network?

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$C$

Show Worked Solution

$\text{Minimum spanning tree:}$

$\text{Using Kruskal’s algorithm:}$

$\text{Edge 1 = 4, Edge 2/3 = 5, Edge 4 = 6, Edge 5 = 9}$

$\therefore\ \text{The largest weighted edge in the MST = 9.}$

$\Rightarrow C$

♦♦ Mean mark 33%.

Networks, STD1 N1 2025 HSC 3 MC

Consider the network diagram.

Which vertex has degree 4?

$A$
$B$
$C$
$D$

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$B$

Show Worked Solution

$\text{Vertex }B\ \text{has 4 edges leading from it so has degree 4.}$

$\Rightarrow B$

Financial Maths, STD1 F1 2024 HSC 27

Zazu works a 38-hour week and is paid at an hourly rate of $45. Any overtime hours worked are paid at time-and-a-half.

In a particular week, Zazu worked the regular 38 hours and some overtime hours. In that week Zazu earned $2790.

How many hours of overtime did Zazu work in that week? (3 marks)

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$\text{16 hours of overtime}$

Show Worked Solution

$\text {Let X = overtime hours}$

	$\text{Total pay}$	$=(38 \times 45)+\Big(X \times \dfrac{3}{2}\times 45\Big) $
	$2790$	$=1710+\dfrac{135}{2}X$
	$\dfrac{135}{2}X$	$=1080$
	$X$	$=\dfrac{1080 \times 2}{135}$
		$=16 \text{ hours of overtime}$

Networks, STD1 N1 2024 HSC 20

The diagram shows a network with weighted edges.

Draw a minimum spanning tree for this network and determine its weight. (2 marks)

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Is it possible to find another spanning tree with the same weight? Give a reason for your answer. (1 mark)

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b. $\text{Yes.}$

$\text{→ The edge}\ FC\ \text{on the MST above could be replaced by the edge}\ BC$

$\text{to create a second MST (with equivalent weight = 24)}$

Show Worked Solution

♦ Mean mark 53%.

b. $\text{Yes.}$

$\text{→ The edge}\ FC\ \text{on the MST above could be replaced by the edge}\ BC$

$\text{to create a second MST (with equivalent weight = 24)}$

♦♦♦ Mean mark 16%.

Algebra, STD1 A1 2019 HSC 34 v1

Given the formula $D=\dfrac{B(x+1)}{18}$, calculate the value of $x$ when $D=90$ and $B=400$. (3 marks)

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$3.05$

Show Worked Solution

$\text{Make}\ x\ \text{the subject:}$

$D$	$=\dfrac{B(x+1)}{18}$
$18D$	$=B(x+1)$
$x+1$	$=\dfrac{18D}{B}$
$x$	$=\dfrac{18D}{B}-1$
$\text{When }$	$D=90, B=400$
$\therefore\ x$	$=\dfrac{18\times 90}{400}-1=3.05$

Financial Maths, STD1 F3 2023 HSC 30

A plumber leases equipment which is valued at $60 000.

The salvage value of the equipment at any time can be calculated using either of the two methods of depreciation shown in the table.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{Method of depreciation} \rule[-1ex]{0pt}{0pt} & \textit{Rate of depreciation} \\
\hline
\rule{0pt}{2.5ex} \text{Straight-line method} \rule[-1ex]{0pt}{0pt} & \text{\$3500 per annum} \\
\hline
\rule{0pt}{2.5ex} \text{Declining-balance method} \rule[-1ex]{0pt}{0pt} & \text{12% per annum} \\
\hline
\end{array}

Under which method of depreciation would the salvage value of the equipment be lower at the end of 3 years? Justify your answer with appropriate mathematical calculations. (3 marks)

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$\text{Straight-line method:}$

$S$	$=V_0-Dn$
	$=60\ 000-3500\times 3$
	$=$49\ 500$

$\text{Declining-balance method:}$

$S$	$=V_0(1-r)^n$
	$=60\ 000(1-0.12)^3$
	$=60\ 000(0.88)^3$
	$=$40\ 888.32$

$\text{Salvage value is lower for the declining-balance method.}$

Show Worked Solution

$\text{Straight-line method:}$

$S$	$=V_0-Dn$
	$=60\ 000-3500\times 3$
	$=$49\ 500$

$\text{Declining-balance method:}$

$S$	$=V_0(1-r)^n$
	$=60\ 000(1-0.12)^3$
	$=60\ 000(0.88)^3$
	$=$40\ 888.32$

$\text{Salvage value is lower for the}$

$\text{declining-balance method.}$

♦ Mean mark 48%.

Measurement, STD1 M3 2023 HSC 29

The diagram shows the location of three places $X$, $Y$ and $C$.

$Y$ is on a bearing of 120° and 15 km from $X$.

$C$ is 40 km from $X$ and lies due west of $Y$.

$P$ lies on the line joining $C$ and $Y$ and is due south of $X$.

Find the distance from $X$ to $P$. (2 marks)

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What is the bearing of $C$ from $X$, to the nearest degree? (2 marks)

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$7.5\ \text{km}$
$259^{\circ}$

Show Worked Solution

a. $\text{In}\ \Delta XPY:$

$\angle PXY=180-120=60^{\circ}$

$\cos 60^{\circ}$	$=\dfrac{XP}{15}$
$XP$	$=15\times \cos 60^{\circ}$
	$=7.5\ \text{km}$

♦♦ Mean mark (a) 24%.

b. $\text{In}\ \Delta XPC:$

$\text{Let}\ \theta = \angle CXP$

$\cos \theta$	$=\dfrac{7.5}{40}$
$\theta$	$=\cos^{-1} \Big(\dfrac{7.5}{40}\Big)$
	$=79.193…$
	$=79^{\circ}\ \text{(nearest degree)}$

$\text{Bearing}\ C\ \text{from}\ X$	$=180+79$
	$=259^{\circ}$

♦♦♦♦ Mean mark (b) 9%.

Algebra, STD1 A3 2023 HSC 26

Electricity provider $A$ charges 25 cents per kilowatt hour (kWh) for electricity, plus a fixed monthly charge of $40.

Complete the table showing Provider $A$ 's monthly charges for different levels of electricity usage. (1 mark)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Electricity used in a month (kWh)} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ 0\ \ \ \ \ & \ \ \ 400\ \ \ & \ \ 1000\ \ \\
\hline
\rule{0pt}{2.5ex} \text{Monthly charge (\$)} \rule[-1ex]{0pt}{0pt} & 40 & & 290 \\
\hline
\end{array}

Provider $B$ charges 35 cents per kWh, with no fixed monthly charge. The graph shows how Provider $B$ 's charges vary with the amount of electricity used in a month.

On the grid on above, graph Provider A's charges from the table in part (a). (1 mark)
Use the two graphs to determine the number of kilowatt hours per month for which Provider $A$ and Provider $B$ charge the same amount. (1 mark)

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A customer uses an average of 800 kWh per month.
Which provider, $A$ or $B$, would be the cheaper option and by how much? (2 marks)

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c. $400\text{ kWh}$

d. $A\text{ is cheaper by }$40.$

Show Worked Solution

♦♦ Mean mark (b) 40%.

c. $\text{Same charge when Provider }A = \text{Provider } B \text{ i.e. where the lines intersect}$

$=400\text{ kWh (see graph above)}$

d. $\text{When kWh}= 800, \ \ A=$240 \text{ and } B=$280$

$\therefore A\text{ is cheaper by }$40.$

Networks, STD1 N1 2023 HSC 18

A network of running tracks connects the points $A, B, C, D, E, F, G, H$, as shown. The number on each edge represents the time, in minutes, that a typical runner should take to run along each track.

Which path could a typical runner take to run from point $A$ to point $D$ in the shortest time? (2 marks)

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A spanning tree of the network above is shown.

Is it a minimum spanning tree? Give a reason for your answer. (2 marks)

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a. $ABFGD$

b. $\text{See worked solutions}$

Show Worked Solution

a. $\text{Using Djikstra’s Algorithm:}$

$\text{Shortest route}$	$=ABFGD$
	$=3+1+5+5$
	$=14$

b. $\text{Total time of given spanning tree}$

$=3+11+1+2+4+5+5$

$=31$

$\text{Consider the MST below:}$

$\text{Total time (MST)}= 3+1+2+4+5+5+9=29$

$\therefore \text{ Given tree is NOT a MST.}$

♦♦ Mean mark (b) 21%.

Networks, STD1 N1 2023 HSC 15

The table shows some of the flight distances (rounded to the nearest 10 km between various Australian cities.

Use the information in the table to complete the network diagram where the edges are labelled with distances. (2 marks)

Mahsa wants to travel from Hobart to Darwin. She wants to change planes only once.
Using the network diagram, calculate how many kilometres she will travel by plane. (1 mark)

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b. $4190\text{ km}$

Show Worked Solution

b. $\text{Changing planes only once}\Rightarrow H\rightarrow S\rightarrow D$

$\text{Kilometres travelled}=1040+3150=4190\text{ km}$

Measurement, STD1 M5 2023 HSC 9 MC

A bag contains 150 jelly beans. Some of them are red and the rest are blue. The ratio of red to blue jelly beans is 2 : 3.

Sophie eats 10 of each colour.

What is the new ratio of red to blue jelly beans?

$2 : 3$
$4 : 9$
$5 : 8$
$11 : 17$

Show Answers Only

$C$

Show Worked Solution

$\text{Original ratio}\ = 2:3$

$\text{5 parts = 150}\ \Rightarrow\ \text{1 part}\ = 150/5=30$

$\text{Original ratio (by number)}\ = 2 \times 30:3 \times 30 = 60:90$

$\text{After eating 10 of each colour:}$

$\text{Ratio}\ = 50:80 = 5:8$

$\Rightarrow C$

♦♦ Mean mark 34%.

Financial Maths, STD1 F1 2023 HSC 7 MC

An item was purchased for a price of $$880$, including $10\%$ GST.

What is the amount of GST included in the price?

$$8.00$
$$8.80$
$$80.00$
$$88.00$

Show Answers Only

$C$

Show Worked Solution

$\text{Let}\ C =\text{Original cost}$

$C+0.1 \times C$	$=880$
$1.1C$	$=880$
$C$	$=\dfrac{880}{1.1}$
	$=$800$

$\therefore \text{GST}=800\times 0.1=$80$

$\Rightarrow C$

♦ Mean mark 15%.

Financial Maths, STD1 F1 2023 HSC 6 MC

A delivery truck was valued at $$65 000$ when new. The value of the truck depreciates at a rate of $22$ cents per kilometre travelled.

What is the value of the truck after it has travelled a total distance of $132 600$ km?

$$35 828$
$$29 172$
$$14 872$
$$14 300$

Show Answers Only

$A$

Show Worked Solution

$\text{Value}$	$=65\ 000-132\ 600\times\frac{22}{100}$
	$=$35\ 828$

$\Rightarrow A$

Financial Maths, STD1 F1 2023 HSC 2 MC

An amount of $2500 is invested at a simple interest rate of 3% per annum.

How much interest is earned in the first two years?

$75
$150
$2575
$2652

Show Answers Only

$B$

Show Worked Solution

$I$	$=Prn$
	$=2500 \times\ \dfrac{3}{100} \times\ 2$
	$=$150$

$\Rightarrow B$

Algebra, STD1 A1 2019 HSC 34

Given the formula `C = (A(y + 1))/24`, calculate the value of `y` when `C = 120` and `A = 500`. (3 marks)

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`4.76`

Show Worked Solution

`text(Make)\ \ y\ \ text(the subject:)`

`C`	`= (A(y + 1))/24`
`24C`	`= A(y + 1)`
`y + 1`	`= (24C)/A`
`y`	`= (24C)/A-1`
	`= (24 xx 120)/500-1`
	`= 4.76`

Financial Maths, STD1 F1 2022 HSC 28

Julie has a gross annual salary of $67 000. During the year she also received an income of $780 from investments and had tax deductions totalling $1000.

The table below shows the income tax rates for the 2021–2022 financial year.

Calculate the tax payable on Julie's taxable income, ignoring the Medicare levy. (3 marks)

Show Answers Only

`$12 170.50`

Show Worked Solution

`text{Taxable Income}`	`=\ text{Total Income – Deductions}`
	`=67\ 000 + 780-1000`
	`=$66\ 780`

`:.\ text{Tax Payable}`	`=5092+0.325(66\ 780-45\ 000)`
	`=5092+7078.50`
	`=$12\ 170.50`

♦ Mean mark 49%.

Networks, STD1 N1 2022 HSC 20

The table below shows the distances, in kilometres, between a number of towns.

Using the vertices given, draw a weighted network diagram to represent the information shown in the table. (2 marks)
A tourist wishes to visit each town.
Draw the minimum spanning tree which will allow for this AND determine its length. (3 marks)

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`1015\ text{km}`

Show Worked Solution

b. `text{Using Prim’s algorithm (starting at}\ Y):`

`text{1st edge:}\ YC`

`text{2nd edge:}\ CB`

`text{3rd edge:}\ SB`

`text{4th edge:}\ YM`

`text{Length of minimum spanning tree}`

`=275 + 150+60+530`

`=1015\ text{km}`

Financial Maths, STD1 F1 2022 HSC 7 MC

Tian is paid $20.45 per hour, as well as a meal allowance of $16.20 per day.

What are Tian's total earnings if she works 9 hours per day for 5 days?

$329.85
$936.45
$1001.25
$1649.25

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`C`

Show Worked Solution

`text{Earnings (5 days)}`	`=5 xx [(9 xx 20.45) + 16.20]`
	`=5 xx 200.25`
	`=$1001.25`

`=>C`

Networks, STD1 N1 2021 HSC 17

The network diagram shows the travel times in minutes along roads connecting a number of different towns.

Draw a minimum spanning tree for this network and determine its length. (3 marks)

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How long does it take to travel from Queentown to Underwood using the fastest route? (1 mark)

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`\text{See Worked Solutions}`
`65 \ \text{minutes}`

Show Worked Solution

a. `text{Using Prim’s algorithm (starting at}\ W):`

♦ Mean mark part (a) 40%.

`text(1st edge:)\ \ WM`

`text(2nd edge:)\ \ WP`

`text(3rd/4th edges:)\ \ MU\ text(and)\ \ WF`

`text(5th edge:)\ \ MK`

`text(6th edge:)\ \ KQ`

`text(7th edge:)\ \ PC`

`text{Length of minimum spanning tree} \ = 160`

b.	`text{Fastest route}\ (Q\ text(to)\ U)`	`= 45 + 20`
		`= 65 \ text{minutes}`

Networks, STD1 N1 2021 HSC 3 MC

Consider the network diagram.

What is the sum of the degrees of all the vertices in this network?

Show Answers Only

`D`

Show Worked Solution

♦♦ Mean mark 31%.

`text{Sum of degrees}`	`= 4 + 3 + 4 + 2 + 3`
	`= 16`

`=> D`

Networks, STD1 N1 2020 HSC 21

The diagram represents a network with weighted edges.

Draw a minimum spanning tree for this network and determine its length. (3 marks)

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The network is revised by adding another vertex, `K`. Edges `AK` and `CK` have weights of 12 and 10 respectively, as shown.

What is the length of the minimum spanning tree for this revised network? (1 mark)

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`text(Length = 14)`

`text(One of many possibilities:)`
`24`

Show Worked Solution

a. `text{Using Kruskal’s Algorithm (one of many possibilities):}`

♦♦ Mean mark part (a) 32%.

`text{Edge 1 :}\ GH\ (1)`
`text{Edge 2 :}\ FH\ (2)`
`text{Edge 3 :}\ CF\ (2)`
`text{Edge 4 :}\ FD\ (2)`
`text{Edge 5 :}\ DE\ (2)`
`text{Edge 6 :}\ BC\ (3)`
`text{Edge 7 :}\ AB\ (2)`

`text{Minimum length of spanning tree}`	`= 1 + 2 + 2 + 2 +2 + 3 +2`
	`= 14`

♦ Mean mark part (b) 45%.

b. `text{Add}\ CK \ text{to the minimum spanning tree in (a).}`

`therefore \ text(Revised length)`	`= 14 + 10`
	`= 24`

Algebra, STD1 A2 2020 HSC 10 MC

A plumber charges a call-out fee of $90 as well as $2 per minute while working.

Suppose the plumber works for `t` hours.

Which equation expresses the amount the plumber charges ($`C`) as a function of time (`t` hours)?

`C = 2 + 90t`
`C = 90 + 2t`
`C = 120 + 90t`
`C = 90 + 120t`

Show Answers Only

`D`

Show Worked Solution

♦♦♦ Mean mark 21%.

`text(Hourly rate)`	`= 60 xx 2`
	`= $120`

`therefore C = 90 + 120t`

Algebra, STD1 A1 2020 HSC 7 MC

The distance between Bricktown and Koala Creek is 75 km. A person travels from Bricktown to Koala Creek at an average speed of 50 km/h.

How long does it take the person to complete the journey?

40 minutes
1 hour 25 minutes
1 hour 30 minutes
1 hour 50 minutes

Show Answers Only

`C`

Show Worked Solution

Mean mark 52%.

`text(Time)`	`= frac(text(Distance))(text(Speed))`
	`= frac(75)(50)`
	`=1.5 \ text(hours)`
	`= 1 \ text(hour) \ 30 \ text(minutes)`

`=> \ C`

Networks, STD1 N1 2019 HSC 28

The network diagram shows the tracks connecting 8 picnic sites in a nature park. The vertices `A` to `H` represents the picnic sites. The weights on the edges represent the distance along the tracks between the picnic sites, in kilometres.

Each picnic site needs to provide drinking water. The main water source is at site `A`.

Draw a minimum spanning tree and calculate the minimum length of water pipes required to supply water to all the sites if the water pipes can only be laid along the tracks. (2 marks)

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One day, the track between `C` and `H` is closed. State the vertices that identify the shortest path from `C` to `E` that avoids the closed track. (1 mark)

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`text(25 km)`
`CGHE`

Show Worked Solution

a. `text(One strategy – using Prim’s algorithm:)`

♦ Mean mark part (a) 45%.

`text(Starting at)\ A`

`text(1st edge -)\ AB,\ \ text(2nd edge -)\ BC`

`text(3rd edge -)\ CH,\ \ text(4th edge -)\ HG`

`text(5th edge -)\ GF,\ \ text(6th edge -)\ HD`

`text(7th edge -)\ DE\ text(or)\ HE`

`text(Maximum length = 4 + 5 + 3 + 2 + 1 + 5 + 5 = 25 km)`

♦♦ Mean mark part (b) 30%.

b. `text(Shortest Path is)\ CGHE`

\(S\)	\(=V_0(1-r)^n\)
\(44\ 000\)	\(=55\ 000(1-r)^1\)
\(\dfrac{44\ 000}{55\ 000}\)	\(=1-r\)
\(1-r\)	\(=0.8\)
\(r\)	\(=1-0.8=0.20\)

\(\text{Total Cost}\ \)	\(\text{= Price + Transfer + Stamp Duty}\)
	\(\text{= }24\ 200+50+726\)
	\(\text{= }$24\ 976\)

\(\text{Loan amount}\ \)	\(\text{= Total Cost + Interest}\)
	\(\text{= }24\ 976+5095.104\)
	\(\text{= }$30\ 071.104\)

\(\text{Monthly repayment}\)	\(=\dfrac{30\ 071.104}{36}\)
	\(=$835.308444\)
	\(\approx $835.31\)

Let	\(P\)	\(\ =\ \text{the number of passenger cars, and}\)
	\(B\)	\(\ =\ \text{the number of motorcycles}\)

\(P(\text{White})\)	\(=\dfrac{50}{100}=\dfrac{1}{2}\)
\(P(\text{Red})\)	\(=\dfrac{25}{100}=\dfrac{1}{4}\)
\(P(\text{Yellow})\)	\(=\dfrac{15}{100}=\dfrac{3}{20}\)
\(P(\text{Green})\)	\(=\dfrac{10}{100}=\dfrac{2}{20}=\dfrac{1}{10}\)

	\(\text{Total pay}\)	\(=(38 \times 45)+\Big(X \times \dfrac{3}{2}\times 45\Big) \)
	\(2790\)	\(=1710+\dfrac{135}{2}X\)
	\(\dfrac{135}{2}X\)	\(=1080\)
	\(X\)	\(=\dfrac{1080 \times 2}{135}\)
		\(=16 \text{ hours of overtime}\)

\(D\)	\(=\dfrac{B(x+1)}{18}\)
\(18D\)	\(=B(x+1)\)
\(x+1\)	\(=\dfrac{18D}{B}\)
\(x\)	\(=\dfrac{18D}{B}-1\)
\(\text{When }\)	\(D=90, B=400\)
\(\therefore\ x\)	\(=\dfrac{18\times 90}{400}-1=3.05\)

\(S\)	\(=V_0-Dn\)
	\(=60\ 000-3500\times 3\)
	\(=$49\ 500\)